Problem 937

Solution. It is known that the position of a plane is determined by three points (not lying on the same line). Let's write the equation of any plane passing through the point $M(1; 2; 0):$

$$A(x-1)+B(y-2)+C z=0$$

To obtain the desired equation of the plane, we need to require that the coordinates of points $N$ and $P$ satisfy equation $(A)$:

$$\left\{\begin{array} { l } { - 3 B + 2 C = 0 , } \\ { - A - B - C = 0 } \end{array} \quad \text { or } \left\{\begin{array}{l} 3 B-2 C=0 \\ A+B+C=0 \end{array}\right.\right.$$

From here, $B=\frac{2}{3} C, A=-C-B=-\frac{5}{3} C$. Substituting these values into equation (A), we get the desired equation of the plane:

$$-\frac{5}{3} C(x-1)+\frac{2}{3} C(y-2)+C z=0$$

or

$$\frac{5}{3}(x-1)-\frac{2}{3}(y-2)-z=0$$

or

$$5 x-2 y-3 z-1=0$$

To determine the angles formed by the normal vector $\bar{n}\{5 ;-2 ;-3\}$ of the desired plane with the coordinate axes, we use formulas (14):

$$\begin{aligned} & \cos \alpha=\frac{5}{\sqrt{25+4+9}}=\frac{5 \sqrt{38}}{38} \approx 0.8111 \\ & \cos \beta=-\frac{2}{\sqrt{38}}=-\frac{\sqrt{38}}{19} \approx-0.3244 \\ & \cos \gamma=-\frac{3}{\sqrt{38}}=-\frac{3 \sqrt{38}}{38} \approx-0.4866 \end{aligned}$$

From here, $\alpha \approx 35^{\circ} 48^{\prime}, \beta \approx 108^{\circ} 56^{\prime}, \gamma \approx 119^{\circ} 7^{\prime}$.